13. Equations with absolute values
Earlier we learned |±a|=a
|a|, also called the "absolute value" of a, returns the value of a after removing its sign.
So, how do we now solve an equation of the form $|x+3|=5$?
One way would be to square both sides, since squaring also effectively removes a negative sign.
(squares of negative and positive numbers are both positive)
This gives: $|x+3|^2=5^2$ or $x^2+6x+9=25$ or $x^2+6x-16=0$
$x^2+6x-16=(x+8)(x-2)=0$, from which we get $x=-8$ or $x=2$
Substituting, we see $|-8+3|=|-5|=5$ and $|2+3|=|5|=5$
e.g. solve: $|x+3|<5$
Squaring both sides, $|x+3|^2<5^2$ or $x^2+6x+9<25$ or $x^2+6x-16<0$
This gives, $x^2+6x-16=(x+8)(x-2)<0$,
which means only one of $(x+8)$ or $(x-2)$ is less than 0 (for their product to be negative)
if $(x+8)<0$, then $x<-8$, and $x-2$ is also negative, which wouldn't work.
so $(x-2)<0$, in which case $x<2$ and $(x+8)>0$ or $x>-8$, which gives us a range of $-8<x<2$ as the desired solution.